CODEWITHSUNDEEP
javaandroid
user2563949
user2563949

Reputation: 161

Java string comparison not working as intended



I've done quite a bit of string comparisons in java in the past, but this time it doesn't seem to work.
I'm aware of the fact, that you have to use the .equals() function in order to compare strings in java.

In my attempt to figure out what's wrong, I wrote the following code snippet:
Log.e("testLogic", String.valueOf(taken.getText().toString().trim().equals('1'))); Log.e("testValue", taken.getText().toString().trim());
producing the following result:

E/testLogic﹕ false
E/testValue﹕ 1
E/testLogic﹕ false
E/testValue﹕ 1
E/testLogic﹕ false
E/testValue﹕ 0

This seems rather strange, since the two first 'testLogic' logs should produce true.
The code is used in a custom list adapter if it means anything.

/Mikkel

Upvotes: 0

Views: 196

Answers (2)

Razib
Razib

Reputation: 11163

The .equals() comes from the Object class. Each class inherit it from the Object class. String class also inherited it from the Object class and override it like this - 

    public boolean equals(Object anObject) {
        if (this == anObject) {
            return true;
        }
        if (anObject instanceof String) {
            String anotherString = (String)anObject;
            int n = count;
            if (n == anotherString.count) {
            char v1[] = value;
            char v2[] = anotherString.value;
            int i = offset;
            int j = anotherString.offset;
            while (n-- != 0) {
                if (v1[i++] != v2[j++])
                return false;
            }
            return true;
            }
        }
        return false;
  }

And String has not any overloaded version of equals method for char. You can see if String's .equals() method takes any object but it returns false when it is any type rather than String. It only returns true if the provided parameter is of type String and is meaningfully equals. So you have to write - 

Log.e("testLogic",String.valueOf(taken.getText().toString().trim().equals("1")));

instead of - 

Log.e("testLogic", String.valueOf(taken.getText().toString().trim().equals('1')));

Upvotes: 0

Francisco Romero
Francisco Romero

Reputation: 13199

It is because you are not comparing 2 Strings. You have to put it like this:

Log.e("testLogic", String.valueOf(taken.getText().toString().trim().equals("1")));

because .equals() function needs two Strings. Supposing that s1 and s2 are Strings, you should do:

s1.equals(s2);

I expect it will be helpful for you!

Upvotes: 2

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